What is the integral of (3^{2y}-27^y)/(9^{2y)} ?

The integral of (3^{2y}-27^y)/(9^{2y)} is -1/(2ln(3)*9^y)+1/(ln(3)*3^y)+C

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integral of (3^{2y}-27^y)/(9^{2y)}

\int \frac{3 ^{2 y} - 2 7 ^{y}}{9 ^{2 y}} d y

- \frac{1}{2 ln ( 3 ) \cdot 9 ^{y}} + \frac{1}{ln ( 3 ) \cdot 3 ^{y}} + C

Solution steps

\int \frac{3 ^{2 y} - 2 7 ^{y}}{9 ^{2 y}} d y

Simplify

\frac{3 ^{2 y} - 2 7 ^{y}}{9 ^{2 y}} : \frac{9 ^{y} - 2 7 ^{y}}{8 1 ^{y}}

= \int \frac{9 ^{y} - 2 7 ^{y}}{8 1 ^{y}} d y

Expand

\frac{9 ^{y} - 2 7 ^{y}}{8 1 ^{y}} : (\frac{1}{9})^{y} - (\frac{1}{3})^{y}

= \int (\frac{1}{9})^{y} - (\frac{1}{3})^{y} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int (\frac{1}{9})^{y} d y - \int (\frac{1}{3})^{y} d y

\int (\frac{1}{9})^{y} d y = - \frac{1}{2 ln ( 3 ) \cdot 9 ^{y}}

\int (\frac{1}{3})^{y} d y = - \frac{1}{ln ( 3 ) \cdot 3 ^{y}}

= - \frac{1}{2 ln ( 3 ) \cdot 9 ^{y}} - (- \frac{1}{ln ( 3 ) \cdot 3 ^{y}})

Simplify

= - \frac{1}{2 ln ( 3 ) \cdot 9 ^{y}} + \frac{1}{ln ( 3 ) \cdot 3 ^{y}}

Add a constant to the solution

= - \frac{1}{2 ln ( 3 ) \cdot 9 ^{y}} + \frac{1}{ln ( 3 ) \cdot 3 ^{y}} + C

\int \frac{1}{3} sin^{3} (x) cos^{3} (x) d x

\int 2 ln (3 x + 1) d x

\int \frac{1}{x ^{2} 9 + 4 x ^{2}} d x

\int \frac{3}{2 - \frac{x}{2}} d x

What is the integral of (3^{2y}-27^y)/(9^{2y)} ?
The integral of (3^{2y}-27^y)/(9^{2y)} is -1/(2ln(3)*9^y)+1/(ln(3)*3^y)+C